If centroid of the tetrahedron , whose vertices are given by (0,0,0), (a, 2, 3),(1, b, 2) and (2, 1, c) be (1, 2, –1), then distance of P(a,b,c) from origin is equal to

\sqrt{107}

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\sqrt{14}

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\frac{\sqrt{107}}{\sqrt{14}}

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10

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Solution

The correct option is A $\sqrt{107}$
Centroid $\equiv (\frac{\sum x}{4}, \frac{\sum y}{4}, \frac{\sum z}{4}) = (1, 2, - 1)$
$\Rightarrow a = 1, b = 5, c = - 9; ∴ \sqrt{a^{2} + b^{2} + c^{2}} = \sqrt{107}$ .

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If centroid of the tetrahedron , whose vertices are given by (0,0,0), (a, 2, 3),(1, b, 2) and (2, 1, c) be (1, 2, –1), then distance of P(a,b,c) from origin is equal to

The correct option is A √107Centroid ≡(∑x4,∑y4,∑z4)=(1,2,−1) ⇒a=1,b=5,c=−9;∴√a2+b2+c2=√107.

The correct option is A $\sqrt{107}$
Centroid $\equiv (\frac{\sum x}{4}, \frac{\sum y}{4}, \frac{\sum z}{4}) = (1, 2, - 1)$
$\Rightarrow a = 1, b = 5, c = - 9; ∴ \sqrt{a^{2} + b^{2} + c^{2}} = \sqrt{107}$ .